2 diagonals of a regular nonagon (a 9-sided polygon) are chosen.  What is the probability that their intersection lies inside the nonagon?
Answer: There are $\binom{9}{2} = 36$ pairs of points in the nonagon, and all but 9 (the sides of the nonagon) are diagonals, which means there are 27 diagonals.  So there are $\binom{27}{2} = 351$ pairs of diagonals.  Any four points on the nonagon uniquely determine a pair of intersecting diagonals. (If vertices $A,B,C,D$ are chosen, where $ABCD$ is a convex quadrilateral, the intersecting pair of diagonals are $AC$ and $BD$.)  So the number of sets of intersecting diagonals is the number of combinations of 4 points, which is $\binom{9}{4} = 126$.  So the probability that a randomly chosen pair of diagonals intersect is $\dfrac{126}{351} = \boxed{\dfrac{14}{39}}$.